链表-代码随想录234. 回文链表、143. 重排链表

链表问题：双指针、迭代法、递归法（天然可用）、虚拟头节点

双链表 既可以向前查询也可以向后查询。

循环链表，顾名思义，就是链表首尾相连。循环链表可以用来解决约瑟夫环问题。

234. 回文链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
数组模拟
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        list=[]
        while head: 
            list.append(head.val)
            head=head.next
        l,r=0, len(list)-1
        while l<=r: 
            if list[l]!=list[r]:
                return False
            l+=1
            r-=1
        return True   
      
#反转后半部分链表
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        fast = slow = head

        # find mid point which including (first) mid point into the first half linked list
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        node = None

        # reverse second half linked list
        while slow:
            # slow.next, slow, node = node, slow.next, slow
            temp = slow.next
            slow.next = node
            node = slow
            slow = temp

        # compare reversed and original half; must maintain reversed linked list is shorter than 1st half
        while node:
            if node.val != head.val:
                return False
            node = node.next
            head = head.next
        return True

 143. 重排链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: Optional[ListNode]) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        d = collections.deque()
        tmp = head
        while tmp.next: # 链表除了首元素全部加入双向队列
            d.append(tmp.next)
            tmp = tmp.next
        tmp = head
        while len(d): # 一后一前加入链表
            tmp.next = d.pop()
            tmp = tmp.next
            if len(d):
                tmp.next = d.popleft()
                tmp = tmp.next
        tmp.next = None # 尾部置空

        return head

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: Optional[ListNode]) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        # d = collections.deque()
        # tmp = head
        # while tmp.next: # 链表除了首元素全部加入双向队列
        #     d.append(tmp.next)
        #     tmp = tmp.next
        # tmp = head
        # while len(d): # 一后一前加入链表
        #     tmp.next = d.pop()
        #     tmp = tmp.next
        #     if len(d):
        #         tmp.next = d.popleft()
        #         tmp = tmp.next
        # tmp.next = None # 尾部置空

        # return head
        if head == None or head.next == None:
            return True
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        right = slow.next # 分割右半边
        slow.next = None # 切断
        right = self.reverseList(right) #反转右半边
        left = head
        # 左半边一定比右半边长, 因此判断右半边即可
        while right:
            curLeft = left.next
            left.next = right
            left = curLeft

            curRight = right.next
            right.next = left
            right = curRight

        return head


    def reverseList(self, head: ListNode) -> ListNode:
        cur = head   
        pre = None
        while(cur!=None):
            temp = cur.next # 保存一下cur的下一个节点
            cur.next = pre # 反转
            pre = cur
            cur = temp
        return pre